An emf of 10 V is applied to a coil with an inductance of 40 mH and a resistance of 0.50 $\varOmega$ .How long does it take for the current to reach one half of its maximum value?
A)0.055 s
B)0.55 s
C)5.5 s
D)55 s

Question

Quizplus · Accepted Answer

The time constant for an LR circuit is given by $\tau = \frac{L}{R}$, where $L$ is the inductance and $R$ is the resistance. Here, $L = 40 \times 10^{-3} H$ and $R = 0.50 \Omega$, so $\tau = \frac{40 \times 10^{-3}}{0.50} = 0.08 s$. The current in an LR circuit reaches $1 - \frac{1}{e}$ (about 63%) of its maximum value after one time constant. To find the time it takes for the current to reach half its maximum value, we use the formula for the growth of current in an LR circuit: $I(t) = I_{max}(1 - e^{-t/\tau})$. Setting $I(t) = 0.5I_{max}$ and solving for $t$ gives $t = -\tau \ln(1 - 0.5) = \tau \ln(2)$. Substituting $\tau = 0.08 s$ yields $t = 0.08 \ln(2) \approx 0.055 s$.

An Emf of 10 V Is Applied to a Coil Ω\varOmegaΩ

An Emf of 10 V Is Applied to a Coil $\varOmega$